The first olympics were held in athens in the year 1896. below is a graph of the gold-medal times for the 100 meter men's freestyle event in swimming since 1896. use the graph below to answer questions 7, 8a, and 8b.
#7a: michael is looking at the data. he says that if you ignore the point (0, 82), the data is pretty linear. he draws the purple line on the graph. what is the meaning of the slope of this line in the context?
#7b: approximately, what is the slope of the purple line?
just enter the number. you do not need to include the units of the slope.
#8a: michael is wondering if it would be reasonable to use his linear model to predict the gold-medal time for the olympics in the year 3000. do you think a linear model is reasonable?
yes, i think that a linear model is reasonable to predict the gold-medal time in the year 3000.
no, i do not think that a linear model is reasonable to predict the gold-medal time in the year 3000.
#8b: explain to michael why you think his model is reasonable or not.

the answer is d. (1.2, -4.7) on edgn

hope this !

step-by-step explanation:

no, her answer is not correct.   we square the lengths of all sides and then determine whether the pythagorean theorem holds here:

9² + 15² = 306.   this is not equal to the square of the longest side ( that square being 395.   answer c is correct.

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.